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## Uniform Contnuity of Real-Valued Functions

### By Maurice Ticas

Uniform continuity is an important concept in real analysis. It's most often introduced after working with the concept of continuity and is a much more stronger statement about a function in the sense that uniform continuity of a real-valued function implies continuity.

What's needed first is the working definition. We say that a function $$f$$ is uniformly continuous on $$A \subset \mathbb{R}$$ whenever we have that $$(\forall \epsilon >0)(\exists \delta >0)(\forall x_1,x_2 \in A)(|x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|< \epsilon)$$ One very straightforward example of a uniformly continuous function is the real-valued identity function $$f$$ with domain $$\mathbb{R}$$. For such a function, we let $$\epsilon >0$$ be arbitrary, $$\delta=\epsilon$$, and $$x_1,x_2 \in \mathbb{R}$$. We then have that $$|f(x_1)-f(x_2)|=|x_1-x_2|<\delta=\epsilon$$ and therefore the function $$f$$ is uniformly continuous.

Our goal will be to prove that $$\frac{1}{x^2+1}$$ is uniformly continuous on $$\mathbb{R}$$, bringing to the reader's attention various ways of writing proof. The next two posts will help you appreciate discovering more than one way to prove a statement. This first post will develop a theoretical toolbox for showing that $$\frac{1}{x^2+1}$$ is uniformly continuous.

Our first tool to introduce is the following statement:

If $$f:[a,\infty) \rightarrow \mathbb{R}$$ is continuous and there exists an $$L \in \mathbb{R}$$ such that $$f(x) \rightarrow L$$ as $$x \rightarrow \infty$$, then $$f$$ is uniformly continuous on $$[a,\infty]$$.

Let $$\epsilon > 0$$. For this specific $$\epsilon$$ we need to show there is a response $$\delta>0$$ such that for any two elements $$x,y \in [a,\infty)$$ within $$\delta$$ apart from each other, $$|f(x)-f(y)|<\epsilon$$. To begin, we know that $$\lim_{x \rightarrow \infty}f(x)=L$$ yields a real number $$N$$ such that for all $$x \geq N$$, $$|f(x)-L|<\frac{\epsilon}{3}$$. This gives way to three cases that need to be considered. The first case is when the two elements $$x,y$$ of the domain are in $$[a,N]$$, the second when they're in $$[N,\infty)$$, and the third when one is in $$[a,N]$$ and the other in $$[N,\infty)$$ not both equal to $$N$$. For each case, it needs to be demonstrated that $$f$$ is uniformly continuous.

The first case follows from knowing that continuous functions on a compact set are uniformly continuous. Hence $$f$$ is uniformly continuous on $$[a,N]$$. This yields the result of there being a $$\delta > 0$$ for which $$|f(x)-f(y)|<\frac{\epsilon}{3}$$ whenever $$|x-y|<\delta$$ for any $$x,y \in [a,N]$$.

For the second case we have that $$x,y\geq N$$ implies $$|f(x)-f(y)|=|f(x)-L+L-f(y)|\leq|f(x)-L|+|f(y)-L|<\epsilon$$ Taking our $$\delta$$ from the first case, it follows that for $$x,y\geq N$$ and $$|x-y|<\delta$$, $$|f(x)-f(y)|<\epsilon$$. This establishes uniform continuity of $$f$$ on $$[N,\infty)$$.

Finally we consider the last case whereby we have without loss of generality that $$x<N<y$$. Using our same $$\delta$$ value from the first case, we then have that for $$|x-y|<\delta$$, $$|f(x)-f(y)|=$$ $$|f(x)-f(N)+f(N)-L+L-f(y)|\leq$$ $$|f(x)-f(N)|+|f(N)-L|+|L-f(y)|<$$ $$\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$ Having shown all three cases, it is now established that for all $$\epsilon >0$$, there is a $$\delta>0$$ such that for all $$x,y \in [a,\infty)$$, if $$|x-y|<\delta$$, then $$|f(x)-f(y)|<\epsilon$$.

Therefore, if $$f$$ is continuous on $$[a,\infty)$$ and $$\lim_{x\rightarrow \infty}f(x)$$ exists and equal to $$L$$, then $$f$$ is uniformly continuous on $$[a,\infty)$$.

Another tool needed has to do with a function being Lipschitz.

We say a real-valued function $$f$$ is Lipschitz if for all $$x \neq y$$ in the domain of $$f$$ we have that $$\left|\frac{f(x)-f(y)}{x-y}\right|\leq M$$ for some bound $$M>0$$.

An additional tool is stated as follows:

If a real-valued function $$f$$ is Lipschitz, then it is uniformly continuous

Let $$\epsilon >0$$ and $$f$$ be defined from $$A \subset \mathbb{R}$$ to $$\mathbb{R}$$. From $$f$$ being Lipschitz, we have that $$\left|\frac{f(x)-f(y)}{x-y}\right|\leq M$$ for all $$x,y \in A$$, where $$M>0$$. That is the same as saying $$\left|f(x)-f(y) \right| \leq M\left|x-y\right|$$. Observe now that $$|f(x)-f(y) |<\epsilon$$ for whenever $$|x-y|<\frac{\epsilon}{M}$$. Hence $$f$$ is uniformly continuous.

The proof for a Lipschitz function being uniformly continuous easily followed from the definition. Proving that an even function uniformly continuous on $$[0,\infty)$$ is also uniformly continuous on $$(-\infty,0]$$ follows easily too from the definition. Indeed, for $$x,y \in [0,\infty)$$, $$|f(-x)-f(-y)|=\left|f(x)-f(y) \right|<\epsilon$$.

One last tool to be used in one of the proofs for $$\frac{1}{x^2+1}$$ being uniformly continuous uses the Mean Value Theorem. We state it now without proof.

If $$f:[a,b]\rightarrow \mathbb{R}$$ is continuous on its domain and differentiable on $$(a,b)$$, then there exists a point $$c \in (a,b)$$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

A follow-up post will follow in a few days with the various proofs for $$\frac{1}{x^2+1}$$ being uniformly continuous.